The operator equation $AB = \lambda BA$ for self-adjoint operators

Suppose that $$A$$ and $$B$$ are self-adjoint bounded linear operators on a Hilbert space and $$\lambda \in \mathbb{C}$$. It turns out that if $$\lambda \notin \{-1, 1\}$$ then $$AB=\lambda BA \implies AB = BA = 0$$.

Does anyone know of any applications of this result?

In the physics context, with $$A$$ and $$B$$ creation operators of two identical particles, the fact that only $$AB=+BA$$ and $$AB=-BA$$ are nontrivially allowed implies that the particles must be either bosons (even under exchange) or fermions (odd under exchange).
$$\require{enclose} \enclose{horizontalstrike}{\style{font-family:inherit;}{\scriptsize\text{Creation operators are not self-adjoint, but still if AB=\lambda BA then AB=\lambda^2 AB hence either \lambda\in\{-1,1\} or AB=0.}}}$$
• For any $q\in (-1,1)$ one can construct operators $A$ and $B$ such that $AB = qBA$, so your last statement is not correct. In the self-adjoint case the condition $AB=\lambda BA$ indeed implies that also $BA = \lambda AB$, hence $AB = \lambda^{2} AB$, but that's not the case in general. May 25, 2020 at 13:13
• Certainly, but it seems to me that your argument as stated would therefore need some qualification as to what types of creation operators $A$ and $B$ you're considering - roughly speaking, "local", as you say. If the operators involve some Wilson lines connecting them to some reference point, one gets into trouble. Evidently, the OP's observation extends beyond self-adjoint operators, but there are restrictions - it would be interesting to understand those. May 25, 2020 at 15:58